A shearing force of 69 N is applied to an aluminum rod with a length of 23 m, a

Question

A shearing force of 69 N is applied to an aluminum rod with a length of 23 m, a cross-sectional area of 2.8 x 10-5 m, and a shear modulus of 52 x 109 N/m2. As a result the rod is sheared
through a distance in mm of A shearing force of 69 N is applied to an aluminum rod with a length of 23 m, a cross-sectional area of 2.8 x 10-5 m, and a shear modulus of 52 x 109 N/m2. As a result the rod is sheared
through a distance in mm of A shearing force of 69 N is applied to an aluminum rod with a length of 23 m, a cross-sectional area of 2.8 x 10-5 m, and a shear modulus of 52 x 109 N/m2. As a result the rod is sheared
through a distance in mm of

Explanation / Answer

F/YA= 69/52*10^9*2.8*10^-5

4.73*10^-5....

change=23*4.73*10^-5=1.089mm

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